NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.6#
Theorem 9: The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc.
So angles in the same segment are equal, and the angle in a semicircle is 90°.
1. In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB? #
Solution
$OA$ and $OB$ are radii, so $OA = OB = 12$ cm and $\triangle OAB$ is isosceles. Its base angles are equal, so:
$$\angle OAB = \angle OBA = \dfrac{180° - 60°}{2}$$$$\angle OAB = \angle OBA = 60°$$
All three angles of $\triangle OAB$ are 60°, so it is equilateral:
$$AB = OA = 12 \text{ cm}$$∴ The chord $AB$ is 12 cm long.
2. Let A and B be two points on a circle with centre O. #
(i) Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?
No. If $X$ and $Y$ are on the circle on the same side of $AB$, they both lie on the same arc, outside the arc $AB$ on the other side. By Theorem 9, each angle is half the angle that this arc $AB$ subtends at the centre:
$$\angle AXB = \dfrac{1}{2}\angle AOB$$$$\angle AYB = \dfrac{1}{2}\angle AOB$$$$\Rightarrow \angle AXB = \angle AYB$$
∴ Such points do not exist: angles in the same segment are always equal.
(ii) Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle?
Not always. If $X$ and $Y$ are on opposite sides of $AB$, the two arcs together make the whole circle, so the two angles at the centre add up to 360°, and the angles at $X$ and $Y$ add up to half of that:
$$\angle AXB + \angle AYB = \dfrac{1}{2} \times 360°$$$$\angle AXB + \angle AYB = 180°$$
So $\angle AXB = \angle AYB$ is possible on opposite sides only when both are 90°, that is, when $AB$ is a diameter.
∴ The statement is true unless $AB$ is a diameter. If $AB$ is a diameter, every point of the circle (on either side) sees $AB$ at 90°.
(iii) If ∠AXB = ∠AYB, and X and Y do not lie on the circle, does the circle through A, B and X also pass through Y?
Yes, provided $X$ and $Y$ are on the same side of $AB$. This is the idea of Theorem 10: if a segment $AB$ subtends equal angles at two points on the same side of it, the four points lie on one circle.
Draw the circle through $A$, $B$ and $X$ (there is exactly one, by Theorem 1). If $Y$ were inside this circle, $\angle AYB$ would be larger than $\angle AXB$; if outside, smaller. Since the angles are equal, $Y$ must be on the circle.
If $X$ and $Y$ are on opposite sides of $AB$, the answer is no in general.
3. Find x in Fig. 5.26. #
Solution
In Fig. 5.26, $A$, $D$, $C$ and $B$ lie on the circle, with $\angle ADC = 100°$ at $D$ and $\angle ABC = x$ at $B$, on opposite sides of $AC$. Let $O$ be the centre.
$\angle ADC$ is subtended at $D$ by the arc $ABC$ (the arc from $A$ to $C$ that does not contain $D$). By Theorem 9, this arc subtends double the angle at the centre:
$$\text{Angle of arc } ABC \text{ at } O = 2 \times 100°$$$$\text{Angle of arc } ABC \text{ at } O = 200°$$
The two arcs from $A$ to $C$ together make the full turn of 360°, so the other arc, $ADC$, subtends:
$$\text{Angle of arc } ADC \text{ at } O = 360° - 200°$$$$\text{Angle of arc } ADC \text{ at } O = 160°$$
$\angle ABC = x$ is subtended at $B$ by the arc $ADC$. By Theorem 9:
$$x = \dfrac{1}{2} \times 160°$$$$x = 80°$$
∴ $x = 80°$. (Notice that $100° + 80° = 180°$ — opposite angles of a cyclic quadrilateral add up to 180°, as Theorem 11 will show.)